Ta có: \(12>9\)

\(6\sqrt{3}>4\sqrt{5}\)

Do đó: \(12+6\sqrt{3}>9+4\sqrt{5}\)

\(\Leftrightarrow\sqrt{12+6\sqrt{3}}>\sqrt{9+4\sqrt{5}}\)

\(x^2=3+5+2\sqrt{15}=8+\sqrt{60}\)

\(y^2=2+6+2\sqrt{12}=8+\sqrt{48}\)

Mà \(60>48\Rightarrow\sqrt{60}>\sqrt{48}\Rightarrow8+\sqrt{10}>8+\sqrt{48}\)

\(\Rightarrow x^2>y^2\Rightarrow x>y\) (do x;y đều dương)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2014}+\sqrt{2015}\)

nên \(\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

theo ket qua cho thay:9.4594<10

Ta có :

\(\sqrt{3}< \sqrt{4}=2\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{24}< \sqrt{25}=5\)

\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)

Vậy ...

\(\sqrt{3}+\sqrt{8}+\sqrt{24}< \sqrt{4}+\sqrt{9}+\sqrt{25}\)

\(=2+3+5=10\)

Vậy: \(\sqrt{3}+\sqrt{8}+\sqrt{24}< 10\)

\(\sqrt{27}-\sqrt{12}-\sqrt{2016}>\sqrt{25}-\sqrt{16}-\sqrt{2025}\)

\(=5-4-45=-44\)

Vậy \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

Có : \(\sqrt{12}< \sqrt{16}=4\)

         \(\sqrt{2016}< \sqrt{2025}\)         => \(\sqrt{12}+\sqrt{2016}< 4+45\)

                                                                 => \(-\sqrt{12}-\sqrt{2016}>-49\)(1)

Lại có : \(\sqrt{27}>\sqrt{25}=5\)(2)

Từ (1),(2) có : \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>5-49\)or \(\sqrt{27}-\sqrt{12}-\sqrt{2016}>-44\)

Bài làm

Ta có: \(28\sqrt{2}\approx39,6\)

           \(\sqrt{14}\approx3,7\)

           \(2\sqrt{147}\approx24,2\)

           \(36\sqrt{4}=72\)

Nên \(36\sqrt{4}>28\sqrt{2}>2\sqrt{147}>\sqrt{14}\left(72>39,6>24,2>3,7\right)\)

Vậy sắp xếp theo thứ tự tăng dần là: \(36\sqrt{4},28\sqrt{2},2\sqrt{147},\sqrt{14}\)

# Học tốt #

\(\sqrt{14}=\sqrt{7}\sqrt{2};2\sqrt{147}=\sqrt{294}\sqrt{2};36\sqrt{4}=\sqrt{2592}\sqrt{2}\)

từ đó so sánh

\(28\sqrt{2}=\sqrt{28^2\cdot2}=\sqrt{784\cdot2}=\sqrt{1568}.\)

\(\sqrt{14}\)

\(2\sqrt{147}=\sqrt{2^2\cdot147}=\sqrt{4\cdot147}=\sqrt{588}\)

\(36\sqrt{4}=\sqrt{36^2\cdot4}=\sqrt{1296\cdot4}=\sqrt{5184}\)

=> Sắp sếp theo thứ tự tăng dần là  \(\sqrt{14};\sqrt{588};\sqrt{1568};\sqrt{5184}\)

=>  \(\sqrt{14};2\sqrt{147};28\sqrt{2};36\sqrt{4}\)

:3

Có \(\sqrt{8}\). 4 = \(\sqrt{\frac{128}{16}}\).4 > \(\sqrt{\frac{81}{16}}\).4 = 9/4 . 4 =9 = 3.3

<=> \(\frac{\sqrt{8}}{3}\)> 3/4 

\(\sqrt{35}+\sqrt{99}< \sqrt{36}+\sqrt{100}=6+10=16\)

Vậy \(\sqrt{35}+\sqrt{99}< 16\)

a) \(9=6+3=6+\sqrt{9}\)

\(6+2\sqrt{2}=6+\sqrt{8}\)

\(\sqrt{8}< \sqrt{9}\) nên \(6+\sqrt{8}=6+2\sqrt{2}< 6+\sqrt{9}=9\)

b) \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}=5+\sqrt{24}\)

\(3^2=9=5+4=5+\sqrt{16}\)

\(\sqrt{16}< \sqrt{24}\Rightarrow3^2< \left(\sqrt{2}+\sqrt{3}\right)^2\Rightarrow3< \sqrt{2}+\sqrt{3}\)

c) \(9+4\sqrt{5}=\left(2+\sqrt{5}\right)^2\)

\(16=\left(2+2\right)^2=\left(2+\sqrt{4}\right)^2\)

\(\sqrt{4}< \sqrt{5}\Rightarrow2+\sqrt{4}< 2+\sqrt{5}\Rightarrow\left(2+\sqrt{4}\right)^2=16< \left(2+\sqrt{5}\right)^2=9+4\sqrt{5}\)

d) \(\left(\sqrt{11}-\sqrt{3}\right)^2=14-2\sqrt{33}=14-\sqrt{132}\)

\(2^2=14-10=14-\sqrt{100}\)

\(\sqrt{100}< \sqrt{132}\Leftrightarrow-\sqrt{100}>-\sqrt{132}\Leftrightarrow14-\sqrt{100}>14-\sqrt{132}\)

\(\Rightarrow2>\sqrt{11}-\sqrt{3}\)